Question 18: Properties of Matter

Question

The increase in the pressure required to decrease the volume ( $\Delta V$ ) of water is $6.3 	imes 10^7 \mathrm{~N} / \mathrm{m}^2$. The percentage decrease in the volume is $\_\_\_\_$ .
(Bulk modulus of water $=2.1 	imes 10^9 \mathrm{~N} / \mathrm{m}^2$.)

Accepted Answer

Use the formula for bulk modulus:
$$ B = -\frac{\Delta P}{\Delta V / V} $$
For decrease in volume, we take only the magnitude:
$$ B = \frac{\Delta P}{\Delta V / V} $$
So,
$$ \frac{\Delta V}{V} = \frac{\Delta P}{B} $$
Given:
$$ \Delta P = 6.3 	imes 10^7 \ 	ext{N/m}^2 $$
$$ B = 2.1 	imes 10^9 \ 	ext{N/m}^2 $$
Substitute:
$$ \frac{\Delta V}{V} = \frac{6.3 	imes 10^7}{2.1 	imes 10^9} $$
$$ \frac{\Delta V}{V} = 3 	imes 10^{-2} = 0.03 $$
Percentage decrease in volume:
$$ 0.03 	imes 100 = 3\% $$
So, the correct answer is:
$$ \boxed{3\%} $$
Option B

Answer

2 %

Answer

3 %

Answer

6 %

Answer

4 %

Properties of Matter

The increase in the pressure required to decrease the volume ( \Delta V ) of water is 6.3 \times 10^7 \mathrm{~N} / \mathrm{m}^2. The percentage decrease in the volume is \_\_\_\_ . (Bulk modulus of water =2.1 \times 10^9 \mathrm{~N} / \mathrm{m}^2.)

Choose the correct answer: